Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $r \neq 0$. $x = \dfrac{r^2 + 7r + 12}{r^2 + 3r} \times \dfrac{r - 8}{2r + 8} $
Solution: First factor the quadratic. $x = \dfrac{(r + 4)(r + 3)}{r^2 + 3r} \times \dfrac{r - 8}{2r + 8} $ Then factor out any other terms. $x = \dfrac{(r + 4)(r + 3)}{r(r + 3)} \times \dfrac{r - 8}{2(r + 4)} $ Then multiply the two numerators and multiply the two denominators. $x = \dfrac{ (r + 4)(r + 3) \times (r - 8) } { r(r + 3) \times 2(r + 4) } $ $x = \dfrac{ (r + 4)(r + 3)(r - 8)}{ 2r(r + 3)(r + 4)} $ Notice that $(r + 3)$ and $(r + 4)$ appear in both the numerator and denominator so we can cancel them. $x = \dfrac{ \cancel{(r + 4)}(r + 3)(r - 8)}{ 2r(r + 3)\cancel{(r + 4)}} $ We are dividing by $r + 4$ , so $r + 4 \neq 0$ Therefore, $r \neq -4$ $x = \dfrac{ \cancel{(r + 4)}\cancel{(r + 3)}(r - 8)}{ 2r\cancel{(r + 3)}\cancel{(r + 4)}} $ We are dividing by $r + 3$ , so $r + 3 \neq 0$ Therefore, $r \neq -3$ $x = \dfrac{r - 8}{2r} ; \space r \neq -4 ; \space r \neq -3 $